MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Maybe this is trivial but lets give it a try anyways..

Obviously there is a forgetful functor from schemes to topological space.. but is it surjective on objects? i.e. I ask whether any topological space is a result of using the forgetful functor on a certain scheme? Certainly this is not true if we consider ONLY affine schemes (which are spectral spaces i.e. Kolmogorov, Compact, compactness preserved upon finite intersection of open compacts, nonempty irreducibles contain a generic point).. but ... hmm wait.. maybe I should require the topological space have the property that each irreducible component must have at least a generic point?

share|cite|improve this question

Both Manny Reyes and wikipedia are slightly off on their statement of Hochster's characterization of spectral spaces (i.e., topological spaces homeomorphic to the spectrum of some commutative ring).

Both are missing the condition that the quasi-compact opens form a base for the topology. A perhaps more transparent characterization of spectral spaces is that they are precisely the inverse limits of finite $T_0$ spaces (see paragraph 2 on the first page of Hochster's paper for both of these). In particular, from the latter characterization it is easy to see that a compact Hausdorff space is spectral iff it is totally disconnected ($\iff$ zero-dimensional $\iff$ Boolean). Thus most compact sober spaces are not spectral, e.g. the closed unit interval is not a spectral space.

Note also that the paper ends with various characterizations of the topological space of a scheme (Proposition 16): these are precisely the spaces homeomorphic to an open subset of a spectral space.

share|cite|improve this answer
I think that Hochster uses the older language of preschemes. In modern terms, he proved that the underlying spaces of separated schemes are exactly the open subsets of spectral spaces (his "locally spectral and quasispectral" spaces) but that the underlying spaces of all schemes are more generally the spaces which are locally spectral. – Andrew Dudzik Mar 26 '12 at 18:59

M. Hochster classified the prime spectra of commutative rings as spectral spaces---quasi-compact T0 spaces whose quasi-compact open subsets are closed under finite intersection and each of whose nonempty irreducible closed subspaces has a unique generic point (see his paper). He additionally classifies the underlying topological spaces of schemes as open subspaces of spectral spaces (Prop. 16 of his paper).

In particular, as you indicate, any irreducible closed subspace of such a space has a unique generic point. (I believe this may even be an exercise in Hartshorne, I'll look for a reference when I get a chance...) So in particular, the two-point space with the indiscrete topology can't be the underlying topological space of a scheme. I'm sure there's a T0 example. I'll edit this if I can think of an example.

share|cite|improve this answer
The Hartshorne ref is Exercise 3.17 in Chapter II and it is on Zariski spaces which are the noetherian topological spaces with unique generic points for their irreducible non-empty closed subsets. The space underlying any noetherian scheme is a Zariski space which is part (a). – Greg Stevenson Oct 27 '09 at 6:07

This doesn't strictly answer the question but is an interesting example: Spec Qbar \otimes_Q Qbar is isomorphic to the Galois group G_Q. Here G_Q has the profinite topology and Spec Qbar \otimes_Q Qbar has the Zariski topology. This is quite a surprise!

Its easy to see this too: Spec Qbar \otimes_Q Qbar is the inverse limit of Spec K \otimes Qbar = Gal(K/Q) (ranging over Galois number fields K).

share|cite|improve this answer
This is the first step in the simplicial resolution of Spec Q as the sheaf quotient of Spec Qbar by G_Q. – S. Carnahan Oct 27 '09 at 6:00

Your idea of requiring that every irreducible closed subset have a generic point (or, equivalently, that you add a generic point to every irreducible closed subset) is called a 'sober space' (and the process of adding a point if it isn't there is called 'soberification'). Wikipedia claims that Hochster proved every compact sober space is Spec of a ring.

share|cite|improve this answer

Sure he has.. and yes I've seen.. but that doesnt answer the question.. I asked for schemes not prime spectra of commutative rings (otherwise the answer is no.. just take for instance any noncompact space)

@Manny: Yep.. I realized that it should at least be sober and T0 .. but your example with indiscrete two point space is certainly not sober ;) ..

any example/nonexample would be great. Right now Im looking at Hausdorff zero-dimensional spaces (not necessarily Stone), but id like to see the general picture too (so question reduces to weather zero dimensional Hausdorff spaces get map surjectively by the forgetful functor on zero dimensional schemes).. oh wait.. all zero dimensional schemes are T2 arent they?

@David: It's not really a surprise. The profinite groups are Stone spaces, a special form of spectral spaces.. so there must be an affine scheme related to them. So you can do the trick for any profinite group using even Boolean rings.

share|cite|improve this answer
Remember that answers don't necessarily appear in order. If you want to refer to another answer, please be explicit. Even better -- use comments to reply to other answers! In particular, you probably should delete this answer and use comments instead. – Scott Morrison Oct 27 '09 at 7:31

@Scott: Unfortunately, I am a bit new here.. so I am not able to insert comments until i gain 50 points, so please bear with this for a while Im about to get those points.

@Peter Clark: I dont think Hochster missed quasicompact opens forms a base for the topology, and I am sure that any spectral space is the spectrum of a commutative ring. In fact, the part with quasicompact opens forming a base is in the definition of spectral spaces, see second paragraph of the paper of Hochster. Wikipedia is the one that missed the part of the definition of spectral space (sober spaces are not enough.. original definition by Hochster never even mentioned sober spaces).

So I'd like to modify things a bit: Is a T0, sober space with quasicompact opens being a base an open subspace of a spectral space (the only thing I removed here is the condition of being compact)? I fear that if you take the alexandroff compactification of a such a space (has all property of spectral but is not compact) then finite intersection of open compacts may not necessarily mean open compact.. I havent yet given this a serious thought. But if this can work with alexandroff compactification, then I have a characterization of spaces that are a result of schemes.

Im doing a new edit.. I think I have proven the following:

A topological space the result of a scheme (by the forgetful functor) iff It is T0, compact opens forms a basis (NOTE: by compact I mean quasicompact and not quasicompact and hausdorff), finite interesections of compact open is compact open, and every irreducible component has a generic point.

=> This side is easy.. since any scheme has this property (last proposition of hochster proves this also.. since any open subspace of a spectral space has this property)

<= if you have such a topological space then take the alexandroff compactification of it. This compactification turns out to be a spectral space (i.e. all the properties of the topological space AND compactness property). The space is an open subspace of its one point compactification which is spectral, thus an open subspace of a spectral space is the topological space that is a result of a scheme (last proposition of Hochster).

share|cite|improve this answer
"@Peter Clark: I dont think Hochster missed quasicompact opens forms a base for the topology." Right. (I never said that he did.) – Pete L. Clark Jul 31 '10 at 15:29

Just another quickie here.. Prop. 16 of Hochsters paper (thanks to Manny for pointing it out) indeed gives us the kind of characterization I need. So for zero dimensional Hausdorff spaces, the answer is immediately YES (just take the Alexandroff compactification of such space and you get a spectral space, and the underlying original Hausdorff zero dimensional space is an open subspace of it)...

Naturally one can try to compactify a T0, sober spaces and ask if it is then a spectral space. Im not yet sure on that, I may post something after my doctor visit.. there might be some problem regarding preservation of compactness upon finite intersection of open compacts (I call them finite intersection property for open compacts)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.